282 BENDING OF RODS. [ART. 29

Expanding the right-hand side we have

We see that the terms on the right-hand side are of the third order of small quantities. We therefore assume as a trial solution 2/ = &cos cnx + Bk* cos Senas ............... (4)

where k is a small quantity analogous to h, and c, B are as yet undetermined constants. Substitute in the differential equation and neglect all powers of k above the third, we then have (1 - c2) n*k cos cnx + (1 - 9c2) JJ&W cos Bcnx = — |n8 (k cos cnx) (FcV sin2 cnx) = — f n*ksc2 (cos cnx — cos 3cm1}. The equation is therefore satisfied if we put

The solution to the third order of small quantities is therefore y = k cos cnx - -$%n*k* cos 3cn# ............... (5)

where c exceeds unity by the small quantity T\?i2F. Let, as before, h represent the distance OC': we have y = h when x = 0, hence

As=fc-&*«» ........................ (6).

Let the lengths of the string and the rod be 2a and 2J, then when # = a, 2/ = 0, and the Z0os£ raZwe of a is given by cna = %7r. We also have

'* ftV) ............ (7)

when terms above the order k8 are neglected. Eliminating a, we

7T 7T

have I = ^r- (1 — -^n2/£2) (1 + |-c'2ft2/6%-) = ^— (1 + •£%n~k*) .. .(8) when the fourth powers of i are neglected.

Since ri* = TiK we have (-==) = ~ (1 + ^ ^-^-). 7 V^U 2Z V 4a2/

Hence T = -rr

when the fourth powers of h/l are rejected. This equation determines the tension necessary to produce a given deflection OG — h.

29. Let us regard the half GB of the bow as a uniform rod having one end C and the tangent at C fixed while the other end B is acted on by a force T whose direction is parallel to the tangent at C.